# How do you sketch the angle whose terminal side in standard position passes through (-8,15) and how do you find sin and cos?

Jun 11, 2018

$\sin t = \frac{15}{17}$
$\cos t = - \frac{8}{17}$

#### Explanation:

The point (x = -8, y = 15), on the angle terminal side, lies in Quadrant 2. Call t the angle.
$\tan t = \frac{y}{x} = - \frac{15}{8}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{225}{64}} = \frac{64}{289}$
$\cos t = \pm \frac{8}{17}$
Because t lies in Quadrant2, --> cos t is negative:
$\cos t = - \frac{8}{17}$
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{64}{189} = \frac{125}{189}$
$\sin t = \pm \frac{15}{17}$
Because t lies in Quadrant 2, --> sin t is positive.
$\sin t = \frac{15}{17}$